Problem: $\dfrac{dy}{dx}=2y^2$ and $y(1)=-1$. $y(3)=$
The differential equation is separable. What does it look like after we separate the variables? $\dfrac{dy}{y^2}=2\,dx$ Let's integrate both sides of the equation. $\int\dfrac{dy}{y^2}=\int2\,dx$ What do we get? $-\dfrac1y=2x+C$ What value of $C$ makes the solution curve pass through the point $(1,-1)$ ? Let's substitute $x=1$ and $y=-1$ into the equation and solve for $C$. $\begin{aligned} -\dfrac1{-1}&=2\cdot1+C\\ \\ \\ 1&=2+C\\ \\ C&=-1 \end{aligned}$ Now use this value of $C$ to find $y$ when $x=3$. $\begin{aligned} -\dfrac1y&=2\cdot 3-1\\ \\ \\ \\ -\dfrac1y&=5\\ \\ \\ y&=-\dfrac15 \end{aligned}$